Plane through a Point and containing a Line:

We can use two values of t, say t = 0 and t = 1, to find points B and C on the line. Then we can use the given point A to obtain the vectors AB and AC, which will lie in the desired plane, P. By forming the cross product, AB ´ AC, we obtain a vector perpendicular to each of AB and AC. This last vector can be used as the normal vector, n, to describe the plane.

First, find the point B, associated with t = 0:

x_B = 2 + 0, y_B = -3 + 0, z_B = 1 + 0.

Hence, we have B(2, -3, 1).

Next, find C using t = 1:

x_C = 2 - 1 = 1, y_C = -3 + 2*1 = -1, z_C = 1 + 4*1 = 5

Hence, we have C(1, -1, 5).

Now determine the two vectors in the plane:

AB = [x_B - x_A]i + [y_B - y_A]j + [z_B - z_A]k

= [2 - (-3)]i + [(-3) - 4]j + [1 - (-1)]k = 5i - 7j + 2k

AC = [x_C - x_A]i + [y_C - y_A]j + [z_C - z_A]k

= [1 - (-3)]i + [(-1) - 4]j + [5 - (-1)]k = 4i - 5j + 6k

Next determine the normal vector from the cross product:

, or

Finally, we use the normal, n, and the point A in the scalar equation of the plane:

a*[x - x_A] + b*[y - y_A] + c*[z - z_A] = -32*[x - (-3)] -22*[y - 4] + 3*[z - (-1)] = 0.

Simplifying, we get

-32x - 22y +3z = 5.

As a check on our work, we can substitute each of the three points into this equation to make sure that they satisfy it:

A(-3, 4, -1): -32*(-3) - 22*4 + 3*(-1) =5

B(2, -3, 1): -32*2 - 22*(-3) + 3*(-1) = 5

C(1, -1, 5): -32*1 - 22*(-1) + 3*(-1) = 5